Question

Evaluate:
Integration(tan^-1 (sec x + tan x) dx),
0 <= x <= (pi/2)​

Answer 1

Given Integrand,

$\displaystyle \sf \int {tan}^{ - 1} (sec \: x \: + tan \: x)dx$

Consider sec x + tan x,

$\sf \: sec \: x + tan \: x \\ \\ \dashrightarrow \sf \: \dfrac{1 + sin \: x}{cos \: x} \\ \\ \dashrightarrow \sf \: \dfrac{cos {}^{2} \dfrac{x}{2} + {sin}^{2} \dfrac{x}{2} + 2sin \dfrac{x}{2}cos \dfrac{x}{2} }{cos {}^{2} \dfrac{x}{2} - {sin}^{2} \dfrac{x}{2} } \\ \\ \dashrightarrow \sf \: \dfrac{(sin \dfrac{x}{2} + cos \dfrac{x}{2} ) {}^{2} }{(sin \dfrac{x}{2} + cos \dfrac{x}{2})( - sin \dfrac{x}{2} + cos \dfrac{x}{2})} \\ \\ \dashrightarrow \sf \: \dfrac{sin \dfrac{x}{2} + cos \dfrac{x}{2}}{cos \dfrac{x}{2} - sin \dfrac{x}{2} }$

Dividing Nr and Dr by tan(x/2),

$\dashrightarrow \sf \: \dfrac{1+tan\dfrac{x}{2}}{ 1- tan \dfrac{x}{2} } \\ \\ \dashrightarrow \sf \: \dfrac{tan \dfrac{\pi}{4} +tan\dfrac{x}{2}}{tan \dfrac{\pi}{4} - tan \dfrac{x}{2} }$

Of the form tan(A + B),

$\dashrightarrow \sf \: tan( \dfrac{x}{2} + \dfrac{\pi}{4} )$

Now,

$\implies \: \displaystyle \sf \int {tan}^{ - 1} \{tan( \dfrac{x}{2} + \dfrac{\pi}{4} ) \}dx \\ \\ \implies \: \displaystyle \sf \int \{ \dfrac{x}{2} + \dfrac{\pi}{4} \}dx \\ \\ \implies \displaystyle \sf \dfrac{1}{2} \int x.dx + \dfrac{\pi}{4} \int dx \\ \\ \implies \boxed{ \boxed{ \sf \dfrac{ {x}^{2} }{4} + \dfrac{\pi x}{4} + C}}$